MHB Combinations Problem - Confirming My Solution

[Sean1]

Hi everyone this is my first post. Please let me know if I am not following the proper etiquette.

I am helping out a friend with a problem. My answer to part (b) does not match the solution my friend has been given.

Four people are selected from a committee of 12 represent the committee at a particular function.
a) How many different groups of four are possible?
12C4 = 495
b) How many groups of four are possible of either the chairperson or the vice chairperson, but not both, must be in the group.
12C4 – 2C2 x 10C2 = 450
Alternatively
1C1 x 1C0 x 10C3 + 1C0 x 1C1 x 10C3 + 1C0 x 1C0 x 10C4
120 + 120 +210 = 450
I think this is correct, but the solution I have been given is 240?

 

[MarkFL]

For both the chairperson and vice-chairperson, we find there are 10 remaining people from which to choose for the other 3 spots on the committee, so we find:

$$N=2{10 \choose 3}=240$$

 

[Sean1]

For both the chairperson and vice-chairperson, we find there are 10 remaining people from which to choose for the other 3 spots on the committee, so we find:

$$N=2{10 \choose 3}=240$$

Many thanks!